Magnetic field due to a ring having $n$ turns at a distance $x$ on its axis is proportional to (if $r$ = radius of ring)
- A$\frac{r}{(x^2 + r^2)}$
- B$\frac{r^2}{(x^2 + r^2)^{3/2}}$
- C$\frac{nr^2}{(x^2 + r^2)^{3/2}}$
- D$\frac{n^2r^2}{(x^2 + r^2)^{3/2}}$
Explore More
Similar Questions
$A$ coil of $50$ turns and $4$ cm radius carries $2$ $A$ current. The magnetic field at its centre is ....... $mT$.
Difficult
View SolutionTwo long parallel wires are at a distance $R$ apart. They carry steady equal currents in the same directions as shown in the figure. The ratio of magnetic fields at $A, B$ and $C$ respectively,is
Medium
View SolutionIn the given diagram,two current-carrying circular loops of radius $R$ and $2R$ are arranged in the $YZ-$ plane and $XZ-$ plane respectively. The common center of both is at the origin $O$. What will be the angle of the resultant magnetic field from the $X-$ axis?
Medium
View Solution$A$ current of $i$ ampere is flowing through each of the bent wires as shown. Find the magnitude of the magnetic field at $O$.
Medium
View SolutionWhat is the ratio of the magnetic field at the center of a circular ring of radius $R$ to the magnetic field at a point on its axis at a distance of $3R$ from the center?
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo